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3 years ago

if the scale on a scale drawing is 1:30, what should you do with each measurement of each drawing, to get an actual dimensions?

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

You would multiply each dimension by 30.

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Tan(x)=2

nydimaria [60]

Tan(x)=2
Or for x=45 tan(x)=1
then for tan(x)=2 ; x must be superior to 45
Then the answer is d. 63.4 degrees

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3 years ago

If I have a garden that's ten yards and I want to plant a plant that requires 2 yards how many plants can I plant

Damm [24]

Pretty sure it's 5. 5X2=10. Sooo

6 0

3 years ago

PLEASE HELP!!!! WILL GIVE THE BRAINLIEST ANSWER!!

Mariana [72]

Answer:

B (Sometimes, if not understood, you can check to see which isn't given or stated already!)

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1 year ago

Joshua is the place kicker for his college football team. Last season he kicked 42 times and never missed. Each field goal score

zalisa [80]

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3 years ago

What is wrong with the following equation?x2+xâˆ’20/xâˆ’4=x+5a. (xâˆ’4)(xâˆ’5)â‰ x2+xâˆ’20b. The left-hand side is not defined f

joja [24]

Answer:

The answer to this question can be defined as follows:

In part (i), the answer is "option d".

In part (ii), the answer is "option 2".

Step-by-step explanation:

Given:

Part (a)

$\Rightarrow \bold{\frac{x^2+x-20}{x-4}=x+5}\\\\$

Solve the above equation:

$\Rightarrow x^2+x-20=(x+5)(x-4)\\\\\Rightarrow x^2+x-20=(x^2-4x+5x-20)\\\\\Rightarrow x^2+x-20=x^2-4x+5x-20\\\\\Rightarrow \boxed{x^2+x-20=x^2+x-20}\\$

Given:

Part (b)

$\Rightarrow \bold{ \lim_{x \to \3} \frac{x^2+x-12}{x-3}= \lim_{x \to 3} (x+4)}\\\\$

Solve the above equation:

factor of $\Rightarrow x^2+x-20 =(x-3)(x+4)$

$\Rightarrow \lim_{x \to \3} \frac{(x-3)(x+4)}{x-3}= \lim_{x \to 3} (x+4)\\\\\Rightarrow \lim_{x \to \3} (x+4)= \lim_{x \to 3} (x+4)\\\\$

apply limit value:

$\Rightarrow (3+4)= (3+4)\\\\\Rightarrow 7= 7$

6 0

4 years ago